# How do you solve x^2-2x-1=x+3?

Apr 13, 2018

$x = 4$
$x = - 1$

#### Explanation:

First, get $0$ on one side.
$\left({x}^{2} - 2 x - 1\right)$ $- \left(x + 3\right)$=$x + 3$ -(x+3)

${x}^{2} - 3 x - 4 = 0$

Next, factor this equation.
$\left(x - 4\right) \left(x + 1\right) = 0$
Therefore,
$\left(x - 4\right) = 0$
$x = 4$
or
$\left(x + 1\right) = 0$,
$x = - 1$

You can verify this using FOIL:
=${x}^{2} + x - 4 x - 4$
=$\left({x}^{2} - 3 x - 4\right)$

Apr 13, 2018

$x = - 1$
$x = 4$

#### Explanation:

${x}^{2} - 2 x - 1 = x + 3$

${x}^{2} - 2 x - 1 - x - 3 = 0$

${x}^{2} - 3 x - 4 = 0$

${x}^{2} - 4 x + x - 4 = 0$

$x \left(x - 4\right) + 1 \left(x - 4\right) = 0$

$\left(x + 1\right) \left(x - 4\right) = 0$

$x + 1 = 0$
$x = - 1$

$x - 4 = 0$
$x = 4$

Apr 13, 2018

$x = - 1 , x = 4$

#### Explanation:

Consider the given equation
${x}^{2} - 2 x - 1 = x + 3$

Subtract $x + 3$ from both sides (Or bring all terms to the left-hand side)

${x}^{2} - 2 x - 1 - x - 3 = x + 3 - x - 3$

Thus,
${x}^{2} - 2 x - 1 - x - 3 = 0$

${x}^{2} - 3 x - 4 = 0$

Split middle term so as to obtain $\left(- 4\right)$ as product and $\left(- 3\right)$ as sum
${x}^{2} + x - 4 x - 4 = 0$

Take common factors out from paired terms
$x \left(x + 1\right) - 4 \left(x + 1\right) = 0$

Then,
$\left(x + 1\right) \left(x - 4\right) = 0$
$x + 1 = 0$ or $x - 4 = 0$
$x = - 1$ or $x = 4$